Thread: n-epsilon relationship for geometric mean limit

Any idea how to find at least one "n" such that:


[ ( x^1/n + y^1/n ) / 2 ]²ⁿ - xy < epsilon


2 <= x <= a, 2 <= y <= a

for any epsilon?


Actually, its well-known that:

lim [ ( x^1/n + y^1/n ) / 2 ]^{2n _{= xy}}

Even for lim n^(1/n) = 1 it is a problem ...

(1/n) (log n) < log(1 + eps) , n > ?? ...